In a consignment of 15 articles, it is found that 3 are defective. If a sample of 5 articles is chosen at…

In a consignment of 15 articles, it is found that 3 are defective. If a sample of 5 articles is chosen at random from it, then the probability of having 2 defective articles is
  1. $\frac{256}{625}$
  2. $\frac{64}{625}$
  3. $\frac{128}{625}$
  4. $\frac{512}{625}$

Solution

Step 1: Define the variables - Total articles $=15$ - Defective articles $=3$ - Non-defective articles $=15-3=12$ - Sample size $=5$ - We need the probability of choosing exactly 2 defective articles in this sample.
Step 2: Use combinations to calculate possible outcomes The probability can be calculated using combinations, as we are dealing with a selection of items without regard to the order. 1. Number of ways to choose 2 defective articles out of 3: $\binom{3}{2}=\frac{3!}{2!(3-2)!}=3$ 2. Number of ways to choose 3 nondefective articles out of 12: $\binom{12}{3}=\frac{12!}{3!(12-3)!}=220$ 3. Total number of ways to choose 5 articles out of 15: $\binom{15}{5}=\frac{15!}{5!(15-5)!}=3003$
Step 3: Calculate the probability The probability of selecting exactly 2 defective articles and 3 non-defective articles is: $\frac{\binom{3}{2} \times\binom{ 12}{3}}{\binom{15}{5}}=\frac{3 \times 220}{3003}=\frac{660}{3003}=\frac{128}{625}$
The probability of having exactly 2 defective articles in the sample is: $\frac{128}{625}$ So, the correct answer is $\frac{128}{625}$.

Asked in: AP EAMCET 2024 (19 May Shift 2)

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